Pure orthorhombic sulphur transforms to stable monoclinic sulphur above 368.5 K. Applying the third law of thermodynamics, the value of entropy (in JK^{-1}) of transformation at 368.5 K is _____ (answer up to two decimal places)

Given: Entropy change associated with heating orthorhombic sulphur from 0 K to 368.5K is 36.86 J K^{-1}.

Entropy change associated with cooling monoclinic sulphur from 368.5 K to 0 K is -37.8 J K^{-1}.

0.94 J/K the problem is done using Hesslaw, by the definition of the third law of thermodynamics entropy at absolute zero is zero

0.94 J/K

See lessthe problem is done using Hesslaw, by the definition of the third law of thermodynamics entropy at absolute zero is zero

Correct answer is 0.94 As per question, we can write ΔS(IV)= ΔS(I) + ΔS(II) + ΔS(III)=0 36.8+ΔS(II)-37.8=0 ΔS(II)=0.94

Correct answer is 0.94

As per question, we can write

ΔS(IV)= ΔS(

I) +ΔS(II) + ΔS(III)=036.8+ΔS(II)-37.8=0

ΔS(II)=0.94

0.94 is the answer.

0.94 is the answer.

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